Irodov 3.405: Eddy Field Extremal on the Orbit – Amit & Deepali Sinha Foundation

Problem Irodov 3.405

In a betatron the growing axially symmetric magnetic field induces an eddy electric field that circles the axis. Using the betatron condition, show that the magnitude of this eddy field, as a function of the distance from the axis, has an extremum on the equilibrium orbit.

A betatron seen along its magnetic field B (drawn out of the page; schematically, its glyphs are denser inside the orbit, where the field is stronger), with an electron on the equilibrium orbit of radius r_0. — A betatron seen along its magnetic field $\vect B$ (drawn out of the page; schematically, its glyphs are denser inside the orbit, where the field is stronger), with an electron on the equilibrium orbit of radius $r_0$ .

Given

axially symmetric $B(r, t)$
orbit radius $r_0$
betatron condition

Find

show that

dE/dr = 0

r = r_0

Solution

Step 1The eddy field at any radius

By symmetry the eddy field at the distance $r$ from the axis is tangent to the circle of radius $r$ and the same all round it. Faraday’s law on that circle gives its magnitude through the flux $\Phi(r)$ the circle encloses:

2\pi r\,E = \frac{\partial\Phi}{\partial t} \quad\Longrightarrow\quad E(r) = \frac{1}{2\pi r}\,\frac{\partial\Phi}{\partial t}, \qquad \Phi(r) = \int_0^r B(r')\,2\pi r'\,dr' \tag{1}

Step 2How the eddy field changes with radius

Writing $\dot\Phi$ for $\partial\Phi/\partial t$ , differentiating (1) with respect to $r$ by the product rule, and using $\partial\dot\Phi/\partial r = 2\pi r\,\dot B(r)$ from the integral:

\frac{dE}{dr} = -\frac{\dot\Phi}{2\pi r^2} + \frac{1}{2\pi r}\cdot 2\pi r\,\dot B(r) = \dot B(r) - \frac{\dot\Phi}{2\pi r^2}

The mean rate of change of the field inside the circle is $\langle\dot B\rangle = \dot\Phi/(\pi r^2)$ , so the second term is half of it:

\frac{dE}{dr} = \dot B(r) - \frac{1}{2}\,\langle\dot B\rangle

(2)

Step 3The derivative on the equilibrium orbit

The betatron condition says that on the orbit the field is half the mean field inside it at every moment. Differentiating it in time at fixed $r_0$ gives the same relation between the rates:

B(r_0) = \frac{1}{2}\,\langle B\rangle \quad\Longrightarrow\quad \dot B(r_0) = \frac{1}{2}\,\langle\dot B\rangle

Putting this into (2) at $r = r_0$ :

\left.\frac{dE}{dr}\right|_{r_0} = \dot B(r_0) - \frac{1}{2}\,\langle\dot B\rangle = 0

(3)

So the eddy field strength is extremal on the equilibrium orbit.

Step 4An example

Take a field that grows with a fixed parabolic profile, $B = (B_0 - ar^2)\,f(t)$ . The flux integral in (1), divided by $2\pi r$ , gives the eddy field:

E = \frac{\dot f}{r}\int_0^r (B_0 - ar'^2)\,r'\,dr' = \dot f\left(\frac{B_0 r}{2} - \frac{a r^3}{4}\right)

It rises, peaks and falls. Setting its derivative to zero locates the peak:

\frac{dE}{dr} = \dot f\left(\frac{B_0}{2} - \frac{3ar^2}{4}\right) = 0 \quad\Longrightarrow\quad r^2 = \frac{2B_0}{3a}

This is the orbit radius that the betatron condition gives for this profile, as the general result requires.

The eddy field against distance from the axis for a parabolic field profile (units $B_0 = a = 1$ ): it peaks at the equilibrium orbit radius $r_0$ .

Check

Each term of (2) is an induction per unit time, which has the units of $E/r$ (V/m per m), as $dE/dr$ must. A uniform field fails the betatron condition at every radius, and indeed then $\dot B = \langle\dot B\rangle$ , so $dE/dr = \dot B/2$ never vanishes: $E = r\dot B/2$ grows steadily with no extremum, and no equilibrium orbit exists. For the parabolic example, the betatron condition applied directly gives the same radius: $B(r_0) = B_0 - ar_0^2$ and $\langle B\rangle/2 = B_0/2 - ar_0^2/4$ are equal when $3ar_0^2/4 = B_0/2$ .

Remark

Whether the extremum is a maximum or a minimum depends on how steeply the field falls. For a field of fixed shape, differentiating (2) once more and using the condition gives $d^2E/dr^2 = (\dot B/r_0)(1 - n)$ on the orbit, with $n = -(r/B)\,dB/dr$ the field index. A betatron built for stable orbits has $0 < n < 1$ , so its electrons ride a minimum of the eddy field; the parabolic profile, with $n = 4$ , puts them on a maximum.